Answer :
The choices experiences colligative properties. These set of properties depend on the amount of solute dissolved in a solvent. These properties include freezing point depression, boiling point elevation, osmotic pressure and vapor pressure lowering. From the choices, the correct answer is A.
Answer : The correct option is, (C) 1.0 M [tex]AlCl_3[/tex]
Explanation :
According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.
1.0 M means that the 1.0 moles of solute present in 1 liter of solution.
Formula used :
[tex]\frac{\Delta p}{p^o}=i\times X_B[/tex]
where,
[tex]p^o[/tex] = vapor pressure of the pure component (water)
[tex]p_s[/tex] = vapor pressure of the solution
[tex]X_B[/tex] = mole fraction of solute
i = Van't Hoff factor
As we know that the vapor pressure depends on the mole fraction of solute and the Van't Hoff factor.
So, the greater the number of particles of solute dissolved the lower the resultant vapor pressure.
(a) The dissociation of 1.0 M [tex]CH_4[/tex] is not possible because it is a non-electrolyte solute. So, the Van't Hoff factor will be, 1.
(b) The dissociation of 1.0 M [tex]NaCl[/tex] will be,
[tex]NaCl\rightarrow Na^++Cl^-[/tex]
So, Van't Hoff factor = Number of solute particles = [tex]Na^++Cl^-[/tex] = 1 + 1 = 2
(c) The dissociation of 1.0 M [tex]AlCl_3[/tex] will be,
[tex]AlCl_3\rightarrow Al^{3+}+3Cl^{-}[/tex]
So, Van't Hoff factor = Number of solute particles = [tex]Al^{3+}+3Cl^{-}[/tex] = 1 + 3 = 4
(c) The dissociation of 1.0 M [tex]BaCl_2[/tex] will be,
[tex]BaCl_2\rightarrow Ba^{2+}+2Cl^{-}[/tex]
So, Van't Hoff factor = Number of solute particles = [tex]Ba^{2+}+2Cl^{-}[/tex] = 1 + 2 = 3
From this we conclude that, 1.0 M [tex]AlCl_3[/tex] has the highest Van't Hoff factor which means that the solution will exhibit the lowest vapor pressure.
Hence, the correct option is, (C) 1.0 M [tex]AlCl_3[/tex]